All categories

3 years ago

HELP!! ILL GIVE BRAINLIEST As we've learned so far, the popular model of the solar system shifted over

Physics

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

geotric earth simulator I think ya I have no idea

You might be interested in

A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?

belka [17]

Answer:

117.72kW

Explanation:

Given data

Mass m= 50kg

height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

WD= force * distance

WD= mgx

WD= 50*9.81*2

WD= 981 Joules

Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

Hence the power spent is 117.72kW

8 0

3 years ago

what is the net force on an object that is experiencing a force of 25 N north, a force of 25 N south, a force of 50 N to the eas

REY [17]

Answer:

5 n

Explanation:

25 and 25 cancel each other out and 50-45 is 5

4 0

3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

In deep space there is very little friction once they are launched into a probe into deep space where there are no external forc

BabaBlast [244]

Continue on the momentum it has. The probe will continue in the same direction it is moving because there are no forces to act against it. I think this is the answer you are looking for...?

3 0

4 years ago

Read 2 more answers

A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in

pantera1 [17]

Answer:

96046 Ns.

Explanation:

We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.

40 km/h in the east

V₁ = 40 i

V₂ = 50j

momentum p₁ = mV₁

= 1500 X 40 i

= 60000 i

Momentum p₂ = mV₂

= 1500 X 50j

= 75000 j

Change in momentum

p₂ - p₁

75000j - 60000i

Magnitude of change

= $\sqrt{(750000)^2 +(60000)^2$

= 96046 Ns.

5 0

3 years ago